1. Find the equation of the axis of symmetry of the graph of $y = (x + 3)^2 - 5$.
2. The x intercepts of the parabola y=2x^2+13x-7
are at (p,0) and (q,0) Find p and q
Enter your answer in the form "p, q ".
3. The x-intercepts of the parabola y=x^2+bx are and Find the equation of the parabola, and submit your answer in y=ax^2+bx+c form.
1. The vertex is (-3, -5)
The axis of symmetry is x = -3
2. To find the x intercepts
2x^2 + 13x - 7 = 0
(2x -1) ( x + 7) = 0
Set both factors to 0 and solve for x and we get that x = 1/2 = q and x = -7 = p
3. Something left out here
The x-intercepts of the parabola y=x^2+bx are ??? and ???
im sorry for number 3 its The x-intercepts of the parabola y=x^2+bx are (-3,0) and (5,0) Find the equation of the parabola, and submit your answer in y=ax^2+bx+c form.
The x coordinate of vertex will occur halfway between the x intercepts = (5 - 3) /2 = 1
And the x coordinate of the vertex = -b / (2a) where a = 1
So
-b / 2 = 1
b = -2
And since (5,0) is on the graphm we can solve for c thusly
0 = 5^2 - 2(5) + c
0 = 15 + c
c = -15
So the equation is
y = x^2 -2x - 15